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A company manufacturers and sells a electric drills per month. The monthly cost and price-demand equations areC(x) = 55000 + 80x,P= 220 - x/30 , 0 < 2 < 5000.(A) Find the production level that results in the maximum profit.Production Level=(B) Find the price that the company should charge for each drill in order to maximize profitPrice=

A company manufacturers and sells a electric drills per month. The monthly cost and-example-1
User Donell
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1 Answer

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10 votes

Solution:

Given:


\begin{gathered} C(x)=55000+80x \\ p=220-(x)/(30) \end{gathered}

To get the production level that will lead to maximum profit, we need to get the revenue.

Revenue (R) is price multiplied by the number of electric drills.

Hence,


\begin{gathered} R=(220-(x)/(30))x \\ R=220x-(x^2)/(30) \end{gathered}

Part A:

Profit is the difference between revenue and cost;


\begin{gathered} Profit=R-C(x) \\ Profit=220x-(x^2)/(30)-(55000+80x) \\ Profit=220x-80x-(x^2)/(30)-55000 \\ Profit=140x-(x^2)/(30)-55000 \end{gathered}

Therefore,


Profit=140x-(x^(2))/(30)-55,000

To get the production level, we find the maximum value from the profit.


\begin{gathered} P(x)=140x-(x^2)/(30)-55,000 \\ Differentiating\text{ the equation,} \\ P^(\prime)(x)=140-(x)/(15) \\ \\ The\text{ maximum occurs when }P^(\prime)(x)=0 \\ 0=140-(x)/(15) \\ (x)/(15)=140 \\ x=15*140 \\ x=2100 \end{gathered}

Therefore, the production level that results in maximum profit is 2100.

Part B:

The price to maximize profit is;


\begin{gathered} p=220-(x)/(30) \\ p=220-(2100)/(30) \\ p=220-70 \\ p=150 \end{gathered}

User Baltasvejas
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